- #1

moxy

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## Homework Statement

Show that the function [itex]f: J → ℝ[/itex] is bounded if [itex]f[/itex] is uniformly continuous on the bounded interval J.

## Homework Equations

J is a bounded interval, so say J = (a,b)

[itex]f[/itex] is uniformly continuous on J, so

[itex]\forall \epsilon > 0[/itex] there exists a [itex]\delta > 0[/itex] such that for [itex]s,t \in J = (a,b)[/itex]

[itex]|f(s) - f(t)| < \epsilon[/itex] whenever [itex]|s - t| < \delta[/itex]

[itex]f: J → ℝ[/itex] is bounded if there exists a real number M such that |f(x)| ≤ M for all x in J.

## The Attempt at a Solution

I think that if I can extend [itex]f[/itex] to the endpoints of J, then I can use the Extreme Value Theorem to say that [itex]f[/itex] attains a min and max value, i.e. is bounded. So I need to define [itex]f(a)[/itex] and [itex]f(b)[/itex].

[itex]f(a) = \lim_{n→∞}{f(a + \frac{1}{n})}[/itex]

[itex]f(b) = \lim_{n→∞}{f(b - \frac{1}{n})}[/itex]

Where clearly [itex]f[/itex] is defined on [itex][a + \frac{1}{n}, b - \frac{1}{n}][/itex] for all [itex]n \in ℕ[/itex]

Am I on the right track? I don't feel like I've used the fact that [itex]f[/itex] is uniformly continuous on J. Is it because [itex]f[/itex] is uniformly continuous that I'm able to define [itex]f(a)[/itex] and [itex]f(b)[/itex]?